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3 years ago

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A hose, of radius 0.018 m, is connected to a water faucet. The water pressure at the point where the hose connects to the fauce

t is 150,000 Pa. The other end of the hose is connected to an outlet of radius 0.011 m, where the water comes out. The total length of the hose and outlet is 15 m.
(a) An experiment shows that the water exists the outlet with a speed 5.2 m/s. What is the flow rate through the hose?
(b) What is the maximum height that the end of the outlet can he held above the faucet, and water still comes out?
(c) Suppose there is a small hole in the hose. How does that affect the speed at which the water comes out? Explain.

Explain each step!

1 answer:

iVinArrow [24]3 years ago

8 0

Answer:

The smaller the hole, the more force.

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A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh

Artyom0805 [142]

Answer:

The electric field is $5.623\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 0.357 m

Charge $Q=5.03\times10^{-6}\ C$

Point charge $q=4.15\times10^{-6}\ C$

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}$

Where, q = point charge

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}$

$E=5.623\times10^{4}\ N/C$

Hence, The electric field is $5.623\times10^{4}\ N/C$

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3 years ago

When light strikes an opaque material, which of the following accurately describes what happens to the light rays? Some of the l

ExtremeBDS [4]

<span>None of the light passes through it; some of the light is absorbed as heat but most is reflected off the surface. This is how you see </span>objects. reflected light from them hits your eye. (Opaque means not transparent)

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4 years ago

Read 2 more answers

What’s 3 times 10 to the 8th power divided by 2.45 times 10 to the 9th power

Virty [35]

0.12244898 is the value obtained when solving the given.

<u>Explanation:</u>

Given:

3 times 10 to the 8th power can be expressed in equation format as $3 \times 10^{8}$ (3 times symbolizes ‘multiplication’ then it is to the tenth power of 8)

2.45 times 10 to the 9th power can be expressed in equation format as $2.45 \times 10^{9}$ (2.45 times symbolizes ‘multiplication’, then it is to the tenth power of 9)

Asked to find the solution when dividing the above,

$\frac{3 \times 10^{8}}{2.45 \times 10^{9}}=\frac{3}{2.45} \times 10^{8-9}$

When the tenth power of any value goes from denominator to numerator and vice-versa, presents in opposite sign to that it possess (like when ‘ $10^{9}$ ' goes to numerator changed as ' $10^{-9}$ ').

Now by solving the above equation, we get

$\frac{3}{2.45} \times 10^{8-9}=1.22 \times 10^{-1}=\frac{1.22}{10}=0.12244898$

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3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

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3 years ago

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

5 0

3 years ago